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JSS3: MATHEMATICS - 1ST TERM

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  1. Binary Number System I | Week 1
    5 Topics
    |
    1 Quiz
  2. Binary Number System II | Week 2
    6 Topics
    |
    1 Quiz
  3. Word Problems I | Week 3
    4 Topics
    |
    1 Quiz
  4. Word Problems with Fractions II | Week 4
    1 Topic
    |
    1 Quiz
  5. Factorization I | Week 5
    4 Topics
    |
    1 Quiz
  6. Factorization II | Week 6
    3 Topics
    |
    1 Quiz
  7. Factorization III | Week 7
    3 Topics
    |
    1 Quiz
  8. Substitution & Change of Subject of Formulae | Week 8
    2 Topics
    |
    1 Quiz
  9. Simple Equations Involving Fractions | Week 9
    3 Topics
    |
    1 Quiz
  10. Word Problems | Week 10
    1 Topic
    |
    1 Quiz



Lesson 1, Topic 3
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Converting from Other Bases to Base 10 Using the Expanded Notation

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Topic Content:

  • Converting from Other Bases to Base 10 Using the Expanded Notation

Worked Example 1.3.1:

Convert the following numbers to base 10 using expanded notation:

a. 107eight
b. 1211three
c. 203six 
d. 4102five
e. 13011four

Solution

a. 107eight

The table shows the place values for 107. Remember that we count 0 from the last digit from right to left. We multiply each digit by the base (which is 8 for this particular question) raised to the power of the place value.

210
107

107eight = 1 × 82 + 0 × 81 + 7 × 8o

= 64 + 0 + 7

= 71ten

∴ 107eight = 71ten

b. 1211three

3210
1211

1211three = 1 × 33 + 2 × 32 + 1 × 31 + 1 × 3o

= 27 + 18 + 3 + 1

=  49ten

∴ 1211three = 49ten

c. 203six 

210
203

203six  = 2 × 62 + 0 × 61 + 3 × 6o

= 72 + 0 + 3

=  75ten

∴ 203six =  75ten

d. 4102five

3210
4102

4102five = 4 × 53 + 1 × 52 + 0 × 51 + 2 × 5o

= 500 + 25 + 0 + 2

=  527ten

∴ 4102five =  527ten

e. 13011four

43210
13011

13011four = 1 × 44 + 3 × 43 + 0 × 42 + 1 × 41 + 1 × 4o

= 256 + 192 + 0 + 4 + 1

=  453ten

∴ 13011four =  453ten

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